AdventOfCode2024/04/index.ts

import fs from "node:fs";
import { printTime, now } from "../util";

const test = false;

const data = fs.readFileSync(
  test ? "./inputs/testinput" : "./inputs/input",
  "utf8"
);

let timer = now.instant();
console.log(
  "part one:",
  data
    .split("\n")
    .slice(0, -1)
    .reduce(
      (coll, curr, index, array) => {
        // 2D string matching is a pain. Let's turn it into a 1D problem!
        // create the columns
        curr.split("").forEach((s, i) => {
          coll[i] += s;
        });
        // create the down-left-diagonals
        curr.split("").forEach((s, i) => {
          coll[i + array.length + index] += s;
        });
        // create the down-right-diagonals
        curr.split("").forEach((s, i) => {
          coll[array.length * 4 + index - i - 2] += s;
        });
        // include the original row
        return [...coll, curr];
      },
      // pre-populate the array so everything's aligned
      Array.from({ length: test ? 10 * 5 - 2 : 140 * 5 - 2 }, () => "")
    )
    .reduce((total, curr) => {
      // and now just do some regexes over the set of 1D strings
      // can't do /XMAS|SMAX/ because overlaps like XMASAMX are allowed!
      let matchesForward = curr.match(/XMAS/g);
      let matchesBackward = curr.match(/SAMX/g);
      if (matchesForward) {
        total += matchesForward.length;
      }
      if (matchesBackward) {
        total += matchesBackward.length;
      }
      return total;
    }, 0),
  printTime(now.instant().since(timer))
);

timer = now.instant();
console.log(
  "part two:",
  data
    .split("\n")
    .slice(0, -1)
    .reduce((coll, curr, index, array) => {
      if (index === 0 || index === array.length - 1) {
        // we can entirely skip the first and last rows, since there's no way to get a cross started
        return coll;
      }
      // look for A's
      return [
        ...coll,
        ...curr.split("").reduce((c, s, i, a) => {
          if (i === 0 || i === a.length - 1) {
            // same as above, edges are useless
            return c;
          }
          // 1 2
          //  A
          // 3 4
          if (s === "A") {
            return [
              ...c,
              array[index - 1][i - 1] +
                array[index - 1][i + 1] +
                array[index + 1][i - 1] +
                array[index + 1][i + 1],
            ];
          }
          return c;
        }, [] as string[]),
      ];
    }, [] as string[])
    .reduce((coll, curr) => {
      // we could have just done this check above,
      // but it's a little more clear, although slower,
      // to iterate again.
      switch (curr) {
        case "MSMS":
        case "SSMM":
        case "SMSM":
        case "MMSS":
          coll += 1;
        default:
          break;
      }
      return coll;
    }, 0),
  printTime(now.instant().since(timer))
);